ma r

Figure 6.2a,b. A soccer player getting ready to kick a soccer ball (a). The free-body diagram of the lower leg is shown in (b).

erted by the quadriceps (see Fig. 6.2). Assuming that the knee stays stationary just before the takeoff, the conservation of angular momentum in the direction perpendicular to the plane of motion is

Thus, using the principle of conservation of angular momentum we are able to make a low estimate of the quadriceps force.

Next let us consider the equation of motion in the vertical direction. The vertical forces acting on the lower leg are the quadriceps force, the weight of the lower leg, and the unknown vertical component of the knee joint force. The resultant of these forces must be equal to the mass of the lower leg times the acceleration of its center of mass in the vertical direction. Using the polar coordinates, presented in Chapter 2 we can show that this acceleration is equal to the square of the angular speed a times the distance r from the center of mass to the rotational axis of the knee.

Fq - 7 kg • 9.81 m/s2 - Fj = 7 kg • (8 rad/s)2 • (0.22 m)

The free-body diagram in Fig. 6.2 shows that Fj pushes against the tibia and therefore is compressive in nature. Note that this force is nearly 50 fold greater than the weight of the lower leg.

Let us next consider the equation of motion of center of mass of the lower leg in the horizontal direction. The only external force acting on the lower leg in the horizontal direction is the knee joint force Hj. This force must be equal to the mass times the acceleration of the center of mass in the horizontal direction. This acceleration component is equal to the angular acceleration a times the distance r:

Equations of motion do not tell us how the total joint force is shared between the various knee ligaments, tibia, and fibula. However, the example shows that the knee sustains high forces during kicking. The knee joint, situated between the body's two longest lever arms, is particularly susceptible to injury.

Example 6.2. Forces Produced by Biceps During High-Speed Forearm Curls. A 25-year-old man knelt facing a table on which his arm rested horizontally in front of his body. He was instructed to press his elbow into the table at all times. Motion of the elbow as the man flexed his forearm as fast as he could were recorded using a high-speed cinema camera (Fig. 6.3). At the instant when the forearm began to rise from the table, angular velocity w was zero and angular acceleration a was 200 rad/s2. Then, when the angle 6 between the table and the forearm increased to 45°, the angular velocity and angular acceleration took on the following values: w = 13 rad/s and a = 100 rad/s2. The man performing the biceps curl weighed 62 kg. The mass of the forearm (including the hand) was 1.36 kg. The radius of gyration p of the forearm about the mass center was 11 cm. The distance L from the center of mass of the forearm to the center of rotation of the elbow was 18 cm. The length D of the forearm was 27 cm. Determine the moment produced by the biceps at the elbow. Determine the force produced by biceps at

Solution: The free-body diagram of the forearm is shown in Fig. 6.3b. The principle of conservation of angular momentum about the fixed point O of the elbow requires that

in which m is the mass of the forearm about the elbow, L is the distance from its center of mass to the elbow, Mb is the moment created at the elbow by the muscles that flex the forearm, Io is the mass moment of inertia of the forearm about the elbow, and a is its angular acceleration. In accord with the parallel axis theorem discussed in Chapter 4, the moment of inertia of the forearm about the elbow Io is given by the relation:

in which Ic denotes the moment of inertia about the mass center of the forearm. The radius of gyration p is related to Ic by the following equation:

Substituting the parameter values given as data into Eqns. 6.4 and 6.5, we find

Thus, at Q = 0, Mb = 14.4 N-m, and at Q = 45°, Mb = 7.7 N-m. (6.6a) Let us next determine the weight the subject must lift statically to generate the same value of muscle moment. Equations of static equilibrium require that

in which m1 is the mass of the weight lifted by the man and D is the length of his forearm. Using Eqns. 6.6a and 6.6b, we find that mi = 4.5 kg at Q = 0°

A typical 25-year-old man in good physical condition can flex his forearm against a resistance of 10 kg; this shows that the biceps force that results from high levels of angular acceleration of the forearm is less than half the maximal force that can be generated by this muscle group. As we shall see later in this chapter, the smaller the force carried by an activated muscle, the faster is its rate of shortening.

A low estimate for biceps force Fb can be obtained by noting that its lever arm d is equal to 4 cm at Q = 0° and 3 cm at Q = 45°. These estimates are Fb = 360 N at Q = 0° and Fb = 257 N at Q = 45°.

For more information on the analysis of elbow forces from high-speed forearm movements, see Amis et al. (1980).

Example 6.3. Inverse Dynamics for Vertical Jumping. Determine the tension in the Achilles tendon during an instant of vertical jump performed by a volleyball player of 75 kg (Fig. 6.4). At the particular instant of the push-off under consideration, his center of mass had a vertical upward acceleration of 12 m/s2. The angle the sole of his feet made with the horizontal axis was 30°. The angular velocity w and angular acceleration a of his feet were measured to be —15 rad/s and —150 rad/s2, respectively. The minus sign indicates that both the angular velocity and angular acceleration were clockwise. The length of the volleyball player's feet was 27 cm, and the length of the toe region of the feet was 7 cm. The moment arm of the Achilles tendon with respect to the center of rotation of the ankle was 4 cm. The foot weighed 1.5 kg. Compute the tensile force FA carried by the Achilles tendon during the instant of push-off described above.

Solution: Considering the free-body diagram of the volleyball player shown in Fig. 6.4a, we find that

in which W is the weight of the volleyball player, FG is the ground force exerted on each foot, and acy is the acceleration of the center of mass of the player in the vertical direction. Substituting the values of the input parameters into Eqn. 6.7a, we find

2 Fg = 75 kg • (12 + 9.81) (m/s2) => Fg = 818 N (6.7b)

If we had neglected the acceleration of the center of mass of the jumper, the ground force would be equal to half the weight (368 N).

Next, let us consider the free-body diagram of the foot. Unlike the whole body, the foot has a small mass, and the forces acting on it, like the ground force, are significantly greater than its mass times acceleration. Thus, although the feet are in motion, the inertial effects are neglected and equations of statics hold. The validity of this assumption can be confirmed by using the equations of dynamics as opposed to statics in determining the force carried by the Achilles tendon. The results not shown here indicate that neglecting the inertia of the foot results in errors of 1% or less in the tendon force.

According to the equations of statics, the net moment acting on the foot at the center of rotation of the ankle must be equal to zero. Thus, when we take the moment of forces acting on the foot with respect to point B we find

-Fa • 0.04 - 1.5 • 9.81 • 0.06 • cos 30° + 818 • (0.035 + 0.16 • cos 30°) = 0

In this equation, we assumed that the distance between the center of mass of the foot and the ankle was 10 cm (see the free-body diagram in Fig. 6.4c). The force carried by the Achilles tendon is almost five times the body weight. Note that the weight of the foot contributed less than 0.3% to the tensile force in the Achilles tendon.

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The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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